The Coriolis “Force” Part 2:
Reading Pond and Pickard Section 6.3 & 8.1-8.5
Ocean Circulation Sections 3.2 , 3.3, 3.4
Once again I recommend checking out these web-sites.
http://www.physics.ohio-state.edu/~dvandom/Edu/newcor.html
http://satftp.soest.hawaii.edu/ocn620/coriolis/
Last lecture ended with the introduction of the fictitious force the Coriolis force. Let’s begin by looking at the magnitude of the acceleration that a particle on the equator feels due to the earth’s rotation. Centrifugal acceleration is equal to v2/R where v is velocity and R is the radius of curvature. First I will deviate from a this with a Question: Why doesn’t this term v2/R appear in the momentum equations that we’ve written before? Answer: Because they were written in Cartesian coordinates. If we were to transform the advective terms in the momentum equation to polar coordinates the v2/R term would appear. So the advective terms in the momentum equation that we’ve been writing in class and the centrifugal acceleration are the same. The both involve a changing velocity vector (Figure 1) and both are nonlinear—they involve velocity squared.
For now we need not worry about the complexity of a fluid and its advective term. How big is the centrifugal acceleration? On the equator the earth spins at 1000 miles/hour or
446 m/s. The earths radius is ~6380 km, so the centrifugal acceleration is (v2/R) is =.03 m/s2. This can be figured out a different way. The angular velocity of a spinning body can be written as , where W is the rotation rate is radians per second (for example if W = 2p the body would rotate once time around – or 2p radians each second). The earth rotates 2p radians every 23 hours 56 minutes and 4 seconds so W= 7.3 x 10-5 s-1 (the Sidereal Day). So the centrifugal acceleration v2/R =. On the equator this acceleration is in the same direction of gravity—but only about 0.3 percent as strong as gravity. Thus for this effect alone a 200 lb person on the pole would with 200.6 lbs on the equator (however there are other process that alter the effective gravity. Fortunately all of them are small and as oceanographers we can live with a constant gravity world-wide ~ 9.8 m2/s).
But that deals with the vertical acceleration. In oceanography and meteorology most of the motion is in the horizontal direction. To discuss how this impacts particles in motion relative to the rotating frame consider the example found in Pond and Pickard (page 42).
A particle begins moving from point A towards point B at velocity v—so the distance x equals vt. However because it is on a rotating plane it ends up at point B’. The distance from B to B’ is the distance that the rotation has deflected the particle. This is equal to the angular velocity (which is equal to Wx=Wvt ) times the time t. So the distance that it moved is Wvt2. The speed of this deflection is the time-derivative of this distance =2Wvt, and the acceleration of this acceleration is the time-derivative of speed =2Wv. Thus the effect of rotation is to deflect a particle to the right (or left depending on the sign of the rotation) at a rate equal to the particle velocity times twice the rotation rate—this is the enigma that the Coriolis frequency (2W) is twice the rotational frequency W. At the pole this rotation rate is W= 7.29 10-5 radians/sec. Thus the Coriolis effect accelerates particles at the North pole to the right at a rate of 2 W times the speed of the particle relative to the rotating earth.
In summary the acceleration = the Coriolis frequency times the normal velocity. Often the letter f is used to symbolize the Coriolis frequency and so this can be written as:
(1)
(2)
If you forget which equation has the negative sign on the right hand remember that in the northern hemisphere the flow turns to the right. So northward velocity (v positive) accelerates the flow to the east (u positive). In contrast a eastward velocity (u positive) accelerates the flow to the south (v negative). In the northern hemisphere f is postive—so equations 1 and 2 describe circular motion in a clockwise direction. Draw this out and see that the flow is always turning to the right and thus flows clockwise. This motion is also called cum-sole meaning with motion moves in the same direction that the sun moves across the sky or anti-cyclonic for it is in the opposite direction of the flow around a cyclone.
The magnitude of Coriolis parameter (which has the units of frequency) changes with
latitude. The Coriolis frequency equals 2Wsin(q), where q is the latitude in degrees and W=2p/T where T is the number of seconds in a Sidereal Day (86164 s). On the north pole f=1.45 x 10 –4, and the inertial period—the time for a particle governed by the
momentum balance described by 1 & 2 to complete one circle, is 11.96 hours. At our latitude f= 9.37 x 10 –5 and the inertial period is 18.6 hours. In the tropics at 18 degrees north f=4.5 37 x 10 –5 and the inertial period is 36 hours. At the equator f=0 and the inertial period is infinity. At 40 degrees south f=- 9.37 x 10 –5 and the Coriolis effect accelerates a moving particle to the left.
If you’re willing to accept at face value these arguments then all you need to know is that f=2 W sin (lat) and that it enters our momentum equation relative to a rotating earth as:
Note that if we multiply equation 1 by u and equation 2 by v (without the sum of the forces term) and add them up we obtain
Or in other words despite the changing motion the kinetic energy remains the same.
A second example is in the atmosphere where we have high pressures and low pressure systems and therefore pressure gradients. In large scale weather systems (100’s-1000’s of km) most of the momentum balance is between the Coriolis force and the pressure gradient (figure 3). In this case winds flow along isobars rather than across them. This type of momentum balance can be written as:
and is called geostrophic flow. It is the first order momentum balance that occurs in much of the atmosphere and the ocean. It is why winds flow clockwise around a high (in the northern hemisphere) and counter-clockwise around a low.
So does the toilet flush the way it does because of the Coriolis effect? Let’s assume that the flow in the toilet is 1 m/s. The Coriolis effect would accelerate the flow at a rate of 1 m/s * 9.37 x 10 –5 m/s2. Thus in the 10 seconds it takes to flush the toilet the fluid moving down the slope of the bowl would acquire a flow to the right at a speed of * 9.37 x 10 –4 m/s2 or less than 1 mm/s. Compare this to the centrifugal accelerations in the toilet (v2/R) (assume R = 20 cm) and we find that the centrifugal accelerations are 5 m/s2—or 40,000 times larger then the Coriols effect. Note that you can compare these two terms
(3)
Where R is the radius of the toilet. Several things to note about equation 3. First for our toilet bowl example it equals over 50,000 -- indicating that effects of the earth’s rotation are over 50,000 times smaller than those associated with centrifugal acceleration. Secondly note that 3 is dimensionless -- so we’ve come up with another non-dimensional number. This one is known as the Rossby Number- after Carl-Gustaf Rossby, (who appeared on the cover of TIME magazine in 1956) More generally this can be written as
(4)
Where L is a characteristic length scale. It is a ratio of the fluid’s inertia to effects of the earth’s rotation. When the Rossby number is near 1 or less effects of rotation are important. (like all non-dimensional numbers it’s value should be viewed as an order of magnitude estimate). Consider a weather system with a length scale of 500 km, winds 20 miles per hour (~ 10 meters/sec) at 40 degrees north. Here the Rossby number is ~ 0.1 and the effects of the earth’s rotation are clearly important. For a Hurricane v=100 miles/hour =44 m/ s The length scale of a hurricane is ~ 200 km. At a latitude of 30 N the Rossby number is about 3—indicating that effects rotation are important but that effects of the winds inertia also play an important role in the dynamics.
In the ocean if currents are 50 cm/s and for f=10-4 any feature with a length scale of less than 20 km will have a Rossby number less than 1. Note that in the ocean because velocities are smaller than in the atmosphere that in the ocean the earth’s rotation effects smaller scale features than in the atmosphere. Subsequently the eddying motion in the ocean is of a much smaller scale than that in the atmosphere. This is a challenge to ocean modelers because to capture all the eddying motion in the ocean requires a much finer resolution than atmospheric modelers require.
In both the atmosphere and the ocean the first order balance is what we call “Geostrophic” in which the pressure gradient is balanced by the Corolis acceleration. In this balance the flow is steady—and thus if you know the pressure gradient you would know the velocity. This has been useful to Oceanographers who can measure the density field in the ocean and calculate the pressure gradient and thus estimate the Geostrophic velocity of the ocean. However, this requires an assumption that the pressure gradient goes to zero at some depth. This can happen if isopycnals tilt in the opposite direction of the barotropic pressure gradient and produce a lower layer with no pressure gradient and thus no geostrophic velocities occurs this level. As such we call this the level the level of no motion. Since we have velocity above the level—and no velocity below—we have vertical shear. Note that this requires a tilting of the isopycnals and this corresponds to a horizontal density gradient.
Thermal Wind Equation
Writing the momentum balance once again for geostrophic flow:
(5 & 6)
Multiplying 5 by r and taking the vertical gradient we get
(7)
If we change the order if differentiation we get
(8)
Now recall the equation for hydrostatic pressure:
(9)
So vertical gradient in Pressure is simply:
(10)
By using equation 10 in 8 we get
(11)
In the ocean the change in density in the vertical is small with respect to changes in density (this is consistent with the Boussinesq approximation) and since f does not change with height 11 can be rewritten as:
(12)
Equations 11 and 12 are known as the thermal wind equations and demonstrate that in a rotating system a horizontal density gradient generates a vertical shear in the flow.
As a simple example imagine if we were able to run our gravitational adjustment problem is a large rotating tank. The surface slope—from the light fluid to the dense fluid would drive the fluid to the right—this can be determined from a geostropic flow for a tilting sea-surface
However, below this level there is a horizontal density gradient and in the case of this set up it produces a total pressure gradient at depth that is equal to the surface pressure gradient- but in the opposite direction. In the non-rotating case this produced a current in the upper layer that flowed in one direction and an equal and opposite current in the lower layer. However, if we were able to rotate the tank the upper layer would flow into the page and the lower layer flow out of the page (Figure 4) Consequently this drives a flow in the opposite direction producing what we call “Thermal Wind Shear” for it’s initial use was to determine atmospheric winds based on temperature measurements.
But we’ve neglected Friction! One place that a level of no motion can be assumed is on the bottom because flow goes to zero at the bottom because of friction—not due to a vanishing pressure gradient. After the exam we will talk about friction, how it is generated and its influence on geophysical flows. Next lecture will be a review.